Timeline for Is there a conjecture about the bounds (constant or a function) of $\sum_{n \le x} \mu(n)/\sqrt{n}$
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17 events
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| Mar 19, 2021 at 17:02 | comment | added | Will Sawin | @PeterHumphries Yes, this is a good point - the terms from the zeroes don't accumulate much over logarithmic ranges. Maybe something funny would happen if you twist by Dirichlet characters somehow, because then there is a chance of having very small $\rho$. | |
| Mar 19, 2021 at 16:59 | comment | added | Peter Humphries | If you look at the explicit formula for $\sum_{n \leq x} \mu(n) n^{-1/2}$, you see terms like $\frac{x^{i\gamma}}{\gamma \zeta'(\rho)}$, whereas for $\sum_{n \leq x} \mu(n)/\sqrt{x}$, you get $\frac{x^{i\gamma}}{\rho \zeta'(\rho)}$. The difference between $\gamma$ and $\rho$ in the denominator isn't significant. | |
| Mar 19, 2021 at 16:58 | comment | added | Peter Humphries | @WillSawin, for Liouville the answer is different and there is a logarithmic main term; see e.g. my honours thesis. For Mobius, there is a bias $1/\zeta(1/2)$, but this is not enough to overcome the oscillations from the zeroes of zeta, as in the usual methods of Chebyshev's bias. | |
| Mar 19, 2021 at 16:54 | comment | added | Will Sawin | @PeterHumphries Are you sure that the transformation between this function and the Mertens function doesn't introduce a logarithmic factor that swamps the $(\log \log \log x)^{5/4}$ term? Say for a random function $\lambda(n)$, $\mathbb E [ ( \sum_{n< X} \lambda(n) / \sqrt{X})^2 ] = 1$ but $\mathbb E[ (\sum_{n < X} \lambda(n)/ \sqrt{n})^2 = \sum_{n <X}1/n \approx \log X$. Non-randomly, we should see larger contributions for each root in the explicit formula for the $\mu(n)/\sqrt{n}$ sum. | |
| Mar 18, 2021 at 0:20 | comment | added | reuns | For $c\ne 0$, $\int_1^\infty (M(x)-cx^{1/2})x^{-s-1}dx= \frac1{s\zeta(s)}-\frac1{c (s-1/2)}$ does have a singularity at $1/2$ so under the RH we can't tell if $M(x)-cx^{1/2}$ changes of sign infinitely often. | |
| Mar 18, 2021 at 0:14 | comment | added | reuns | If $\sum_{n\le x} a_n$ doesn't change of sign infinitely often then $F(s)=\int_1^\infty (\sum_{n\le x}a_n)x^{-s-1}dx$ has a singularity at $\sigma$ its abscissa of convergence. $a(n)=\mu(n)/n$ gives $F(s)=1/(s\zeta(s+1))$, no singularity on $[-1/2,0]$ whence $\sum_{n\le x} \mu(n)/n$ changes of sign infinitely often. $1/\zeta(1/2)-\sum_{n\le x} \mu(n)n^{-1/2}$ gives $F(s)=\frac1{s\zeta(1/2+s)}-\frac1{s\zeta(1/2)}$, no singularity on $[0,1/2]$ whence $1/\zeta(1/2)-\sum_{n\le x} \mu(n) n^{-1/2}$ changes of sign infinitely often. | |
| Mar 18, 2021 at 0:06 | history | edited | Shree | CC BY-SA 4.0 |
corrected typo
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| Mar 18, 2021 at 0:04 | comment | added | Shree | IMO $\sum_{n<x} {\mu(n)}/{n}$ changing sign infinitely many times happens because its mean/median happens to be $0$. It is equivalent to $\sum_{1<n<x} {\mu(n)}/{n}$ crossing $-1$ infinitely times without ever changing sign, and $\sum_{n<x} {\mu(n)}/{\sqrt{n}}$ crossing $1/\zeta(1/2)$ infinitely many times. | |
| Mar 17, 2021 at 22:13 | history | edited | Shree | CC BY-SA 4.0 |
edited body
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| Mar 17, 2021 at 22:07 | history | edited | Shree | CC BY-SA 4.0 |
Added numerically computed data, in case anyone is interested in details
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| Mar 17, 2021 at 21:54 | comment | added | Peter Humphries | I don't believe that it's known that it changes sign infinitely often, but it should be proveable by current computational methods. I spoke about this with Tim Trudgian a year and a half ago, though I don't know if he's done any computations on it since. | |
| Mar 17, 2021 at 20:28 | comment | added | Shree | @PeterHumphries, Thanks. One more q: is it already known that the partial sum changes sign often, or is it a conjecture? I was wondering if the integral part added to $M(x)/\sqrt{x}$ to get the partial sum of interest may counterbalance the shift to positive to keep the mean and median near $-0.684765...$ and thus keep it negative for a very long time (e.g. $x>10^{10^{39}}$). | |
| Mar 17, 2021 at 20:06 | comment | added | Peter Humphries | Also, there is a bias in this partial sum; assuming standard conjectures, $\sum_{n \leq x} \mu(n) n^{-1/2}$ should have a limiting logarithmic distribution whose median and mean is equal to $1/\zeta(1/2) \approx -0.684765\ldots$. Nonetheless, it still changes sign infinitely often. | |
| Mar 17, 2021 at 20:04 | comment | added | Peter Humphries | As @reuns mentions, it should be the same as for the Mertens function except divided through by $\sqrt{x}$. In particular, the natural conjecture is that $$0 < \limsup_{x \to \infty} (\log \log \log x)^{-5/4} \sum_{n \leq x} \mu(n) n^{-1/2} < \infty$$ and $$-\infty < \liminf_{x \to \infty} (\log \log \log x)^{-5/4} \sum_{n \leq x} \mu(n) n^{-1/2} < 0.$$ See also my answer here: mathoverflow.net/a/368511/3803 | |
| Mar 17, 2021 at 19:56 | comment | added | reuns | This is really the same as the Mertens conjecture in the form $M(x)=O(x^{1/2})$. Wikipedia says that there is a conjecture that it is false with $M(x)\approx O(x^{1/2}(\log\log \log x)^a) $ for some $a>0$. | |
| Mar 17, 2021 at 19:55 | history | edited | Shree | CC BY-SA 4.0 |
added 362 characters in body
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| Mar 17, 2021 at 19:33 | history | asked | Shree | CC BY-SA 4.0 |